Question

`int_0^(pi/2)sin^3x=`?

Answer 1

The definite integral is `2/3`
`0.6666666666666667`

Explanation:

`int_(0)^(pi/2) sin^3x`
Now,
`=int(-cos^2x) sin(x)dx`
Substitute,
`color(grey)(u=cos(x)->dx=-((1)/(sinx))dx`
Therefore,
`=int(u^2-1)dx`
Applying linearity,
`=int u^2du-int 1du` ---(1)

Now, applying power rule in,
`int u^2du`
`=u^3/3` ---(2)
Now, applying constant rule in,
`=int 1du`
`=u` ---(3)
Putting (2) and (3) in (1),
`intu^2du-int1du`
`=u^3/3-u`
Undo substitution in,
`color(grey)(u=cos(x))`
Therefore,
`=(cos^3x)/(3)-cosx`
`therefore =cos^3x/(3)-cosx+C`